JEE Main 2 April 2026 Shift 1: Questions and Solutions

Answer: Option (3)

Key idea: Quadratic Equations with Series Summation and Vieta's Formulas

1. Expand the general term: $(x+k)(x+k+2) = x^2 + (2k+2)x + k^2 + 2k$ for $k = 0$ to $n-1$
2. Sum all terms using summation formulas: $\sum_{k=0}^{n-1}k = \frac{n(n-1)}{2}$ and $\sum_{k=0}^{n-1}k^2 = \frac{n(n-1)(2n-1)}{6}$
3. Simplify to get: $x^2 + (n+1)x + \frac{2n^2 + 3n - 29}{6} = 0$
4. Apply Vieta's formulas: Sum of roots $2\alpha + 2 = -(n+1)$ gives $\alpha = \frac{-n-3}{2}$
5. Use product of roots: $\alpha(\alpha+2) = \frac{2n^2 + 3n - 29}{6}$ and substitute $\alpha$ to get $n^2 = 49$, so $n = 7$
6. Calculate $\alpha = \frac{-7-3}{2} = -5$, therefore $n + \alpha = 7 + (-5) = 2$

Answer: Option (4)

Key idea: Complex number operations: rationalization and equating real and imaginary parts

1. Step 1: Rationalize $\frac{2x}{1+3i}$ by multiplying by $\frac{1-3i}{1-3i}$ to get $\frac{x(1-3i)}{5}$
2. Step 2: Rationalize $\frac{y}{1-2i}$ by multiplying by $\frac{1+2i}{1+2i}$ to get $\frac{y(1+2i)}{5}$
3. Step 3: Substitute and simplify: $50 \cdot \frac{1}{5}[x(1-3i) - y(1+2i)] = 31 + 17i$ gives $10[(x-y) + i(-3x-2y)] = 31 + 17i$
4. Step 4: Equate real parts: $10(x-y) = 31$, so $x - y = 3.1$
5. Step 5: Equate imaginary parts: $10(-3x-2y) = 17$, so $-3x - 2y = 1.7$
6. Step 6: Solve the system: From $x = y + 3.1$ and $-3(y+3.1) - 2y = 1.7$, get $y = -2.2$ and $x = 0.9$
7. Step 7: Calculate $10(x - 3y) = 10(0.9 - 3(-2.2)) = 10(7.5) = 75$

Answer: Option (2)

Key idea: Condition for inconsistent system of linear equations using determinants and row reduction

1. Write the augmented matrix and apply row operations: $R_2 \to R_2 - 2R_1$ and $R_3 \to R_3 - 8R_1$
2. Apply $R_3 \to R_3 - 4R_2$ to get the matrix in row echelon form
3. The last row becomes $[0 \quad 0 \quad \beta-4\alpha \mid -2]$
4. For no solution, we need $\beta - 4\alpha = 0$ (rank of coefficient matrix $< $ rank of augmented matrix) and $-2 \neq 0$ (which is satisfied)
5. From $\beta - 4\alpha = 0$, we get $\beta = 4\alpha$
6. Therefore, $\frac{\beta}{\alpha} = 4$

Answer: Option (2)

Key idea: Matrix equations, determinants, and properties of adjoint matrices

1. Find $\alpha$ from $A^2 - 4A + I = O$: Calculate $A^2$ and solve to get $\alpha = 3$
2. Find $\beta$ from $B^2 - 5B - 6I = O$: Calculate $B^2$ and solve to get $\beta = 4$
3. Check (S1): Calculate $(B-A)(B+A) = \begin{bmatrix}13 & 15\\ 7 & 10\end{bmatrix}$, then transpose gives $\begin{bmatrix}13 & 7\\ 15 & 10\end{bmatrix} \neq \begin{bmatrix}13 & 15\\ 7 & 10\end{bmatrix}$, so (S1) is FALSE
4. Check (S2): Calculate $\det(A+B) = \det\begin{bmatrix}4 & 5\\ 5 & 5\end{bmatrix} = -5$
5. Use property $\det(\text{adj}(M)) = [\det(M)]^{n-1}$ for $n \times n$ matrix. For $n=2$: $\det(\text{adj}(A+B)) = \det(A+B) = -5$, so (S2) is TRUE
6. Conclusion: Only (S2) is correct

Answer: Option (2)

Key idea: Intersection of two arithmetic progressions and divisibility conditions

1. Step 1: Write general terms: Set $A$ has $a_n = 5n - 4$ for $n = 1$ to $101$, Set $B$ has $b_m = 7m + 2$ for $m = 1$ to $71$
2. Step 2: For $A \cap B$, solve $5n - 4 = 7m + 2$, which gives $5n = 7m + 6$. For integer $n$, we need $7m \equiv -1 \pmod{5}$, so $m \equiv 2 \pmod{5}$
3. Step 3: Values of $m$ are $2, 7, 12, 17, ..., 67$ (since $m \leq 71$ and corresponding $n \leq 101$). This gives 14 common elements
4. Step 4: Common elements form an A.P.: $16, 51, 86, 121, 156, 191, 226, 261, 296, 331, 366, 401, 436, 471$ with first term $16$ and common difference $35$
5. Step 5: For divisibility by 3: $16 + 35k \equiv 0 \pmod{3}$ gives $1 + 2k \equiv 0 \pmod{3}$, so $k \equiv 1 \pmod{3}$
6. Step 6: Values $k = 1, 4, 7, 10, 13$ give elements $51, 156, 261, 366, 471$ (all divisible by 3)
7. Step 7: Count the elements: There are $\boxed{5}$ elements in $A \cap B$ divisible by 3

Answer: Option (3)

Key idea: Permutations with repetition - distributing objects with constraints

1. We have 5 distinct digits {1, 2, 3, 5, 7} and need to form 7-digit numbers with each digit used at least once. Since $5 + 2 = 7$, we need 2 extra digits.
2. **Case 1:** Two digits appear twice each, three appear once. Choose 2 digits to repeat: $\binom{5}{2} = 10$ ways.
3. Arrange 7 digits with two pairs: $\frac{7!}{2! \times 2!} = \frac{5040}{4} = 1260$. Case 1 total: $10 \times 1260 = 12600$.
4. **Case 2:** One digit appears three times, four appear once. Choose 1 digit to repeat thrice: $\binom{5}{1} = 5$ ways.
5. Arrange 7 digits with one triplet: $\frac{7!}{3!} = \frac{5040}{6} = 840$. Case 2 total: $5 \times 840 = 4200$.
6. Total number of 7-digit numbers: $12600 + 4200 = 16800$.

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Key idea: Binomial Coefficients and Relations

Answer: Option (3)

Key idea: Mean of grouped data and finding equation from roots

1. Find class marks: $x_i = 7.5, 12.5, 17.5, 22.5, 27.5, 32.5$
2. Calculate total frequency: $\Sigma f_i = 2 + k + 28 + 54 + (k+1) + 5 = 90 + 2k$
3. Calculate $\Sigma f_i x_i = 2(7.5) + k(12.5) + 28(17.5) + 54(22.5) + (k+1)(27.5) + 5(32.5) = 1910 + 40k$
4. Use mean formula: $21 = \frac{1910 + 40k}{90 + 2k}$
5. Solve: $21(90 + 2k) = 1910 + 40k \Rightarrow 1890 + 42k = 1910 + 40k \Rightarrow k = 10$
6. Check which equation has $x = 10$ as root: $2(10)^2 - 19(10) - 10 = 200 - 190 - 10 = 0$ ✓

Answer: Option (3)

Key idea: Finding vertices from midpoints and calculating incentre coordinates

1. Step 1: Use midpoint formulas to set up equations: If midpoints are $D(\frac{5}{2}, 7)$, $E(\frac{5}{2}, 3)$, $F(4, 5)$, then $x_2 + x_3 = 5$, $x_1 + x_3 = 5$, $x_1 + x_2 = 8$ (and similar for $y$-coordinates).
2. Step 2: Solve the system by adding all equations: $2(x_1 + x_2 + x_3) = 18 \Rightarrow x_1 + x_2 + x_3 = 9$. Similarly, $y_1 + y_2 + y_3 = 15$.
3. Step 3: Find vertices: $A(4, 1)$, $B(4, 9)$, $C(1, 5)$ by subtracting appropriate equations.
4. Step 4: Calculate side lengths: $a = BC = 5$, $b = AC = 5$, $c = AB = 8$.
5. Step 5: Apply incentre formula: $h = \frac{ax_A + bx_B + cx_C}{a+b+c} = \frac{5(4) + 5(4) + 8(1)}{18} = \frac{8}{3}$ and $k = \frac{5(1) + 5(9) + 8(5)}{18} = 5$.
6. Step 6: Calculate $3h + k = 3 \times \frac{8}{3} + 5 = 8 + 5 = 13$.

Answer: $\frac{8\sqrt{5}}{3}$

Key idea: Ellipse with major axis along y-axis, eccentricity relation, and latus rectum formula

1. Since $a < b$, the major axis is along y-axis. Use $e = \frac{c}{b}$ where $c^2 = b^2 - a^2$
2. From $e = \frac{\sqrt{5}}{3}$: $c = \frac{\sqrt{5}b}{3}$, so $\frac{5b^2}{9} = b^2 - a^2$, giving $a^2 = \frac{4b^2}{9}$
3. Substitute point $(4,3)$ into ellipse equation: $\frac{16}{a^2} + \frac{9}{b^2} = 1$
4. Replace $a^2 = \frac{4b^2}{9}$: $\frac{36}{b^2} + \frac{9}{b^2} = 1$, so $b^2 = 45$ and $a^2 = 20$
5. Use latus rectum formula $L = \frac{2a^2}{b} = \frac{40}{\sqrt{45}} = \frac{40}{3\sqrt{5}} = \frac{8\sqrt{5}}{3}$

Answer: Option (1)

Key idea: Product of sines at specific angles and compound angle formulas

1. Step 1: Recognize that $\sin\left(\frac{\pi}{18}\right)\sin\left(\frac{5\pi}{18}\right)\sin\left(\frac{7\pi}{18}\right) = \sin(10°)\sin(50°)\sin(70°)$
2. Step 2: Use the identity $\sin(x)\sin(60°-x)\sin(60°+x) = \frac{1}{4}\sin(3x)$ with $x=10°$ to get $K = \frac{1}{4}\sin(30°) = \frac{1}{8}$
3. Step 3: Calculate $\frac{10K\pi}{3} = \frac{10 \cdot \frac{1}{8} \cdot \pi}{3} = \frac{5\pi}{12}$
4. Step 4: Express $\sin\left(\frac{5\pi}{12}\right) = \sin(75°) = \sin(45° + 30°)$
5. Step 5: Apply sum formula: $\sin(45°)\cos(30°) + \cos(45°)\sin(30°) = \frac{\sqrt{2}}{2} \cdot \frac{\sqrt{3}}{2} + \frac{\sqrt{2}}{2} \cdot \frac{1}{2} = \frac{\sqrt{6}+\sqrt{2}}{4}$
6. Step 6: Simplify $\frac{\sqrt{6}+\sqrt{2}}{4} = \frac{\sqrt{2}(\sqrt{3}+1)}{4} = \frac{\sqrt{3}+1}{2\sqrt{2}}$

Answer: Option (4)

Key idea: Trigonometric equations with integer constraints and range analysis

1. Step 1: Expand $\sin x(\sin x + \cos x) = \sin^2 x + \sin x \cos x$
2. Step 2: Use identities to get $\frac{1 - \cos 2x}{2} + \frac{\sin 2x}{2} = \frac{1 + \sin 2x - \cos 2x}{2}$
3. Step 3: Simplify to $\frac{1 + \sqrt{2}\sin(2x - \pi/4)}{2} = a$, which gives $\sqrt{2}\sin(2x - \pi/4) = 2a - 1$
4. Step 4: Since $-\sqrt{2} \leq 2a - 1 \leq \sqrt{2}$, we get $a \in \{0, 1\}$
5. Step 5: For $a = 0$: $\sin(2x - \pi/4) = -\frac{1}{\sqrt{2}}$ gives $x \in \{-\pi, -\pi/4, 0, 3\pi/4, \pi\}$ (5 values)
6. Step 6: For $a = 1$: $\sin(2x - \pi/4) = \frac{1}{\sqrt{2}}$ gives $x \in \{-3\pi/4, -\pi/2, \pi/4, \pi/2\}$ (4 values)
7. Step 7: Total $n(S) = 5 + 4 = 9$

Answer: Option (3)

Key idea: Intersection of lines in 3D space and condition for point on xy-plane

1. Parametrize both lines: Line 1 with parameter $t$ gives $(3t-1, 5t-a, 7t-b-1)$ and Line 2 with parameter $s$ gives $(s+2, 4s+b, 7s+2a)$
2. Equate coordinates at intersection: From $x$: $3t - s = 3$; From $y$: $5t - 4s = a + b$; From $z$: $7(t-s) = 2a + b + 1$
3. Since intersection lies on $xy$-plane, $z = 0$. Using Line 1: $7t - b - 1 = 0 \Rightarrow t = \frac{b+1}{7}$
4. Solve for $s$ from equation 1: $s = 3t - 3$. Substitute into the $y$ equation to get $-7t + 12 = a + b$
5. Substitute $t = \frac{b+1}{7}$ into $-7t + 12 = a + b$ to get $11 = a + 2b$
6. Use the $z$ equation to get another relation: $2a = 18 - 3b$. Solving both equations gives $a = 3, b = 4$
7. Therefore, $a + b = 3 + 4 = 7$

Answer: Option (3)

Key idea: Vector Magnitude Optimization using Triangle Inequality and Calculus

1. Express $|3\vec{a} + 2\vec{b}|^2 = 72(1 + \cos\theta)$ and $|3\vec{a} - 2\vec{b}|^2 = 72(1 - \cos\theta)$ where $\theta$ is the angle between $\vec{a}$ and $\vec{b}$
2. Substitute to get $f(\theta) = 18\sqrt{2(1 + \cos\theta)} + 24\sqrt{2(1 - \cos\theta)}$
3. Let $x = \cos\theta$ and find critical points: $f'(x) = \frac{9\sqrt{2}}{\sqrt{1+x}} - \frac{12\sqrt{2}}{\sqrt{1-x}} = 0$
4. Solve to get $9\sqrt{1-x} = 12\sqrt{1+x}$, which gives $\cos\theta = -\frac{7}{25}$
5. Calculate $f(-\frac{7}{25}) = 18 \cdot \frac{6}{5} + 24 \cdot \frac{8}{5} = \frac{108 + 192}{5} = 60$
6. Verify this exceeds boundary values at $\theta = 0$ and $\theta = \pi$

Answer: Option (3)

Key idea: Foot of perpendicular from a point to a line in 3D, Cross product for direction vector

1. Find direction vector of line $L$ using cross product: $\vec{d} = (2\hat{i} + 2\hat{j} + \hat{k}) \times (\hat{i} + 2\hat{j} + 2\hat{k}) = 2\hat{i} - 3\hat{j} + 2\hat{k}$
2. Write equation of line $L$ passing through $(1,1,1)$: $\vec{r} = (1,1,1) + t(2,-3,2)$
3. Point $P$ on line $L$ is $(1+2t, 1-3t, 1+2t)$. Since $\vec{OP} \perp \vec{d}$, we have $\vec{OP} \cdot \vec{d} = 0$
4. Solve: $2(1+2t) - 3(1-3t) + 2(1+2t) = 0 \Rightarrow 1 + 17t = 0 \Rightarrow t = -\frac{1}{17}$
5. Find $P$: $a = \frac{15}{17}$, $b = \frac{20}{17}$, $c = \frac{15}{17}$, so $a+b+c = \frac{50}{17}$
6. Calculate: $34(a+b+c) = 34 \times \frac{50}{17} = 100$

Answer: Option (2)

Key idea: Limits with indeterminate forms, Taylor series expansion, and parameter determination

1. For the limit to exist, both numerator and denominator must approach 0 as $x \to 2$. This gives $2a + b = 12$.
2. Substitute $u = x - 2$ and expand $x^3 - 5x^2 + ax + b = (a-8)u + u^2 + u^3 + (2a+b-12)$.
3. For a non-zero finite limit, the coefficient of $u$ must be zero: $a - 8 = 0$, so $a = 8$ and $b = -4$.
4. Using Taylor expansions: $\sqrt{x-1} - 1 \approx \frac{u}{2}$ and $\ln(x-1) \approx u$ as $u \to 0$.
5. The limit becomes $\lim_{u \to 0} \frac{\sin(u^2(1+u))}{\frac{u^2}{2}} = \lim_{u \to 0} \frac{u^2(1+u)}{\frac{u^2}{2}} = 2$, so $m = 2$.
6. Therefore, $a + b + m = 8 + (-4) + 2 = 6$.

Answer: Option (3)

Key idea: First-order linear differential equation with variable separable method and integration by parts

1. Separate variables: $\frac{dy}{y} = (2 + \ln x)dx$
2. Integrate both sides: $\ln y = \int(2 + \ln x)dx$
3. Use integration by parts for $\int \ln x\, dx = x\ln x - x$, giving $\ln y = 2x + x\ln x - x + C = x(1 + \ln x) + C$
4. Apply initial condition $(1, e)$: $\ln e = 1(1 + \ln 1) + C \Rightarrow 1 = 1 + C \Rightarrow C = 0$
5. Particular solution: $\ln y = x(1 + \ln x)$
6. Find $f(e)$: $\ln y = e(1 + \ln e) = e(1 + 1) = 2e$, so $y = e^{2e}$

Answer: Option (2)

Key idea: Critical points of piecewise functions involving absolute values

1. 1. Check continuity: $\lim_{x \to 0} \sqrt{\frac{|\sin x|}{x}} = 1 = f(0)$, so $f$ is continuous everywhere
2. 2. For $x \neq 0$, write $f(x) = \sqrt{\frac{|\sin x|}{x}}$ and note that $|\sin x| = 0$ at $x = \pm\pi$ in the given interval
3. 3. At $x = \pm\pi$: Since $\sin(\pm\pi) = 0$, we have $f(\pm\pi) = 0$ and the derivative doesn't exist (cusps)
4. 4. Check $x = 0$: Using L'Hôpital or Taylor series, $f'(0^+) = f'(0^-) = 0$, so $x = 0$ is a critical point
5. 5. For other points, differentiate in regions where $\sin x$ has constant sign. Critical points occur when $x\cos x = \sin x$ (i.e., $\tan x = x$) or $x\cos x = -\sin x$
6. 6. The equation $\tan x = x$ has no solutions in $(-2\pi, 2\pi) \setminus \{0\}$ within the valid regions
7. 7. Total critical points: $x = -\pi, 0, \pi$ giving **3 critical points**

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Answer: Option (4)

Key idea: First-order linear differential equation solved by variable separation

1. Rearrange and separate variables: $\frac{dy}{y+1} = \frac{-\cos x}{1+\sin x}dx$
2. Integrate both sides: $\ln|y+1| = -\ln|1+\sin x| + C$
3. Simplify to get: $(y+1)(1+\sin x) = K$
4. Apply initial condition $y(0)=0$ to find $K=1$, giving $(y+1)(1+\sin x) = 1$
5. Solve for $y$: $y = \frac{-\sin x}{1+\sin x}$
6. Substitute $y = -\frac{1}{2}$: $\frac{-\sin \alpha}{1+\sin \alpha} = -\frac{1}{2}$
7. Solve to get $\sin \alpha = 1$, therefore $\alpha = \frac{\pi}{2}$

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Key idea: Domain of composite functions involving logarithm and square root with rational expressions

Answer: 91

Key idea: Telescoping series and sum of squares formula

1. Given $\sum_{k=1}^{6} a_k = 6n^3$, recognize this as a telescoping sum where $a_k = 6(k^3 - (k-1)^3)$
2. Expand: $a_k = 6(3k^2 - 3k + 1) = 18k^2 - 18k + 6$
3. Find the difference: $a_{k+1} - a_k = 18(k+1)^2 - 18(k+1) + 6 - 18k^2 + 18k - 6 = 36k$
4. Substitute into the expression: $\frac{a_{k+1} - a_k}{36} = \frac{36k}{36} = k$
5. Evaluate: $\sum_{k=1}^{6} k^2 = \frac{6 \cdot 7 \cdot 13}{6} = 91$

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Key idea: Quadratic inequalities and probability - condition for quadratic to be always positive using discriminant

Answer: 8

Key idea: Circle passing through origin with equal intercepts on axes, chord length formula

1. Since the circle intersects axes at exactly 3 points with equal intercepts, it passes through $(0,0)$, $(a,0)$, and $(0,a)$ for some $a > 0$
2. The center equidistant from these three points is at $(\frac{a}{2}, \frac{a}{2})$ (can be verified by setting distances equal)
3. The radius squared is $r^2 = (\frac{a}{2})^2 + (\frac{a}{2})^2 = \frac{a^2}{2}$
4. Distance from center $(\frac{a}{2}, \frac{a}{2})$ to line $x+y=1$ is $d = \frac{|a-1|}{\sqrt{2}}$
5. Using chord length formula: $(\sqrt{14})^2 = 4(r^2 - d^2)$, we get $14 = 4(\frac{a^2}{2} - \frac{(a-1)^2}{2})$
6. Simplifying: $14 = 2(2a-1)$ gives $a = 4$
7. Therefore, $r^2 = \frac{16}{2} = 8$

Answer: 192

Key idea: Definite Integration using substitution and King's property

1. For $I_1 = \int_0^{2\sqrt{3}} \log_2(x^2+4)\,dx$, substitute $x = 2\tan\theta$ to get $I_1 = \int_0^{\pi/3} \log_2(4\sec^2\theta) \cdot 2\sec^2\theta\,d\theta$
2. Simplify: $I_1 = 4\int_0^{\pi/3} \sec^2\theta\,d\theta + 4\int_0^{\pi/3} \log_2(\sec\theta) \cdot \sec^2\theta\,d\theta = 4\sqrt{3} + 4J$
3. For the second part $J$, substitute $u = \tan\theta$ and use integration by parts to get $J = \frac{1}{2\ln 2}[u\ln(1+u^2) - 2u + 2\arctan(u)]_0^{\sqrt{3}} = \sqrt{3} - \frac{\sqrt{3}}{\ln 2} + \frac{\pi}{3\ln 2}$
4. Thus $I_1 = 8\sqrt{3} - \frac{4\sqrt{3}}{\ln 2} + \frac{4\pi}{3\ln 2}$
5. For $I_2 = \int_2^4 \sqrt{2^x-4}\,dx$, substitute $2^x = 4\sec^2\phi$ to get $I_2 = \frac{4}{\ln 2}\int_0^{\pi/3} \tan^2\phi\,d\phi = \frac{4\sqrt{3}}{\ln 2} - \frac{4\pi}{3\ln 2}$
6. Adding: $\alpha = I_1 + I_2 = 8\sqrt{3}$ (the extra terms cancel)
7. Therefore $\alpha^2 = (8\sqrt{3})^2 = 64 \times 3 = 192$

Answer: Option (2)

Key idea: Dimensional Analysis of Energy Density

1. Find dimensions of electric field: $[E] = \frac{[F]}{[q]} = \frac{[MLT^{-2}]}{[AT]} = [MLT^{-3}A^{-1}]$
2. Find dimensions of $\varepsilon_0$ from Coulomb's law: $[\varepsilon_0] = \frac{[q^2]}{[F][L^2]} = [M^{-1}L^{-3}T^4A^2]$
3. Calculate $[\varepsilon_0 E^2] = [M^{-1}L^{-3}T^4A^2] \times [MLT^{-3}A^{-1}]^2 = [M^{-1}L^{-3}T^4A^2] \times [M^2L^2T^{-6}A^{-2}]$
4. Simplify: $[\varepsilon_0 E^2] = [M^{1}L^{-1}T^{-2}]$ (This is energy density!)
5. Identify: $a = 1$, $b = -1$, $c = -2$
6. Calculate: $2a - b + c = 2(1) - (-1) + (-2) = 2 + 1 - 2 = 1$

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Key idea: Unknown - question content not provided

Answer: Option (2)

Key idea: Acceleration from velocity field using convective derivative

1. Given velocity field: $\vec{v} = -x\hat{i} + 2y\hat{j} - z\hat{k}$ m/s
2. Use convective derivative: $\vec{a} = \frac{d\vec{v}}{dt} = \frac{\partial \vec{v}}{\partial x}v_x + \frac{\partial \vec{v}}{\partial y}v_y + \frac{\partial \vec{v}}{\partial z}v_z$
3. Calculate: $\vec{a} = (-\hat{i})(-x) + (2\hat{j})(2y) + (-\hat{k})(-z) = x\hat{i} + 4y\hat{j} + z\hat{k}$
4. At point (1, 2, 4): $\vec{a} = \hat{i} + 8\hat{j} + 4\hat{k}$ m/s²
5. Magnitude: $|\vec{a}| = \sqrt{1^2 + 8^2 + 4^2} = \sqrt{81} = 9$ m/s²

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Key idea: Kinematics, Linear Momentum, Force, Angular Momentum, and Torque in 2D motion

Answer: Option (2)

Key idea: Kepler's Third Law of Planetary Motion

1. Apply Kepler's Third Law: $T^2 = \frac{4\pi^2}{GM} R^3$, where $M$ is the mass of the central star
2. For planet $P_1$: $T_1^2 = \frac{4\pi^2}{G(2M)} \times R^3 = \frac{2\pi^2 R^3}{GM}$
3. For planet $P_2$: $T_2^2 = \frac{4\pi^2}{G(4M)} \times (2R)^3 = \frac{4\pi^2 \times 8R^3}{4GM} = \frac{8\pi^2 R^3}{GM}$
4. Find the ratio: $\frac{T_2^2}{T_1^2} = \frac{8\pi^2 R^3/GM}{2\pi^2 R^3/GM} = \frac{8}{2} = 4$
5. Therefore: $\frac{T_2}{T_1} = \sqrt{4} = 2$

Answer: Option (3)

Key idea: Angular acceleration as second derivative of angular displacement

1. Given: $\theta = \frac{5t^4}{40} - \frac{t^3}{3} = \frac{t^4}{8} - \frac{t^3}{3}$
2. Find angular velocity: $\omega = \frac{d\theta}{dt} = \frac{4t^3}{8} - \frac{3t^2}{3} = \frac{t^3}{2} - t^2$
3. Find angular acceleration: $\alpha = \frac{d\omega}{dt} = \frac{3t^2}{2} - 2t$
4. Substitute $t = 10$ s: $\alpha = \frac{3(10)^2}{2} - 2(10) = 150 - 20$
5. Therefore, $\alpha = 130$ rad/s²

Answer: Option (2)

Key idea: Capacitors in series with dielectric; Capacitance of parallel plate capacitor

1. Original capacitance: $C = \frac{\varepsilon_0 A}{d}$
2. With half-filled dielectric, treat as two capacitors in series: Air gap ($d/2$) and dielectric gap ($d/2$)
3. Air capacitor: $C_1 = \frac{\varepsilon_0 A}{d/2} = 2C$; Dielectric capacitor: $C_2 = \frac{K\varepsilon_0 A}{d/2} = 10C$
4. For series combination: $\frac{1}{C_{new}} = \frac{1}{2C} + \frac{1}{10C} = \frac{6}{10C}$, so $C_{new} = \frac{5C}{3}$
5. Percentage increase $= \frac{C_{new} - C}{C} \times 100\% = \frac{5C/3 - C}{C} \times 100\% = \frac{2}{3} \times 100\% = 66.67\%$

Answer: Option (4)

Key idea: First Law of Thermodynamics for Isobaric Process (Diatomic Gas)

1. For a diatomic gas: $C_v = \frac{5}{2}R$ and $C_p = \frac{7}{2}R$
2. At constant pressure: $\Delta Q = nC_p\Delta T = n\frac{7}{2}R\Delta T$
3. Change in internal energy: $\Delta U = nC_v\Delta T = n\frac{5}{2}R\Delta T$
4. Work done at constant pressure: $\Delta W = nR\Delta T$
5. Taking ratio: $\Delta Q : \Delta U : \Delta W = \frac{7}{2} : \frac{5}{2} : 1 = 7:5:2$
6. Verify using First Law: $\Delta Q = \Delta U + \Delta W \Rightarrow 7 = 5 + 2$ ✓

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Key idea: Electric potential and field for conducting spheres in equilibrium

Answer: Option (4)

Key idea: Wave velocity from wave equation

1. Write the given wave equation: $y = 5 \cos \pi \left(200t - \frac{x}{150}\right)$
2. Expand to standard form: $y = 5 \cos \left(200\pi t - \frac{\pi x}{150}\right)$
3. Compare with $y = A \cos(\omega t - kx)$ to get $\omega = 200\pi$ rad/s and $k = \frac{\pi}{150}$ rad/cm
4. Calculate wave velocity: $v = \frac{\omega}{k} = \frac{200\pi}{\pi/150} = 200 \times 150 = 30000$ cm/s
5. Convert to m/s: $v = \frac{30000}{100} = 300$ m/s

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Key idea: Electric field due to dipole at axial and equatorial positions

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Key idea: Zener Diode Voltage Clipping Circuit - Back-to-Back Configuration

Answer: Option (1)

Key idea: Electromagnetic Induction: Power dissipation in coils with varying parameters (turns, area, wire radius)

1. Induced EMF in a coil: $\varepsilon = NA\frac{dB}{dt}$. For coil 2: $\varepsilon_2 = 2N \cdot 2A \cdot \frac{dB}{dt} = 4\varepsilon_1$
2. Resistance of coil wire: $R = \rho \frac{L}{A_{wire}}$ where $L = N \times \text{perimeter} \propto N\sqrt{A}$ and $A_{wire} = \pi r^2$
3. For coil 1: $R_1 \propto \frac{N\sqrt{A}}{r^2}$. For coil 2: $R_2 \propto \frac{2N\sqrt{2A}}{(3r)^2} = \frac{2\sqrt{2}N\sqrt{A}}{9r^2}$
4. Ratio: $\frac{R_2}{R_1} = \frac{2\sqrt{2}}{9}$
5. Power dissipated: $P = \frac{\varepsilon^2}{R}$, so $\frac{P_2}{P_1} = \left(\frac{\varepsilon_2}{\varepsilon_1}\right)^2 \cdot \frac{R_1}{R_2} = 16 \times \frac{9}{2\sqrt{2}} = \frac{72}{\sqrt{2}} = 36\sqrt{2}$
6. Given $P_2 = \sqrt{2}\alpha P_1$, we get $\sqrt{2}\alpha = 36\sqrt{2}$, therefore $\alpha = 36$

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Answer: Option (2)

Key idea: Thin prism at minimum deviation - relationship between incident angle and minimum deviation

1. For a thin prism with refractive index $\mu = 1.5$, the minimum deviation is $\delta_m = A(\mu - 1) = A(1.5 - 1) = 0.5A$
2. At minimum deviation for a symmetric prism, the incident angle $i$ and emergent angle $e$ are equal: $i = e$
3. Using the prism formula at minimum deviation: $i + e = A + \delta_m$, which gives $2i = A + \delta_m$
4. Substituting $\delta_m = 0.5A$: $2i = A + 0.5A = 1.5A$, so $i = 0.75A$
5. The ratio is $\frac{i}{\delta_m} = \frac{0.75A}{0.5A} = \frac{3}{2}$
6. Therefore, $i : \delta_m = 3:2$

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Key idea: Unknown - question text not provided

Answer: Option (2)

Key idea: Photoelectric Effect - Einstein's Equation and Threshold Wavelength

1. Write Einstein's photoelectric equation for both cases: $\frac{hc}{\lambda} = \phi + 3eV_0$ and $\frac{hc}{2\lambda} = \phi + eV_0$
2. Subtract the second equation from the first: $\frac{hc}{\lambda} - \frac{hc}{2\lambda} = 2eV_0$
3. Simplify to get: $\frac{hc}{2\lambda} = 2eV_0$, therefore $hc = 4eV_0\lambda$
4. Substitute back into second equation: $2eV_0 = \phi + eV_0$, so work function $\phi = eV_0$
5. At threshold wavelength: $\frac{hc}{\alpha\lambda} = \phi$, substitute values: $\frac{4eV_0\lambda}{\alpha\lambda} = eV_0$
6. Solve for $\alpha$: $\frac{4}{\alpha} = 1$, therefore $\alpha = 4$

Answer: Option (1)

Key idea: Forces on charged particle in electromagnetic wave; Relation between E and B in EM waves

1. From the given equation, electric field amplitude is $E_0 = 300$ V/m
2. For an EM wave, magnetic field amplitude is $B_0 = E_0/c = 300/(3 \times 10^8) = 10^{-6}$ T
3. Maximum electric force on electron: $F_{E,max} = eE_0 = 300e$ N
4. Since electron moves in y-direction with speed $v = 1.5 \times 10^6$ m/s and $\vec{B}$ is perpendicular, maximum magnetic force: $F_{B,max} = evB_0 = e \times 1.5 \times 10^6 \times 10^{-6} = 1.5e$ N
5. Ratio: $\frac{F_{E,max}}{F_{B,max}} = \frac{E_0}{vB_0} = \frac{300}{1.5} = 200$

Answer: Option (3)

Key idea: Bohr's Model of Hydrogen Atom - Quantization of Angular Momentum and Energy Levels

1. Given angular momentum $L = \frac{3h}{\pi}$
2. Use Bohr's quantization condition: $L = \frac{nh}{2\pi}$
3. Equate: $\frac{3h}{\pi} = \frac{nh}{2\pi}$ which gives $n = 6$
4. Apply energy formula: $E_n = -\frac{13.6}{n^2}$ eV
5. Calculate: $E_6 = -\frac{13.6}{36} = -0.38$ eV

Answer: Option (2)

Key idea: Surface Energy and Surface Tension

1. Given: Initial radius $R = 1$ mm $= 10^{-3}$ m, $n = 512$ droplets, surface tension $T = 0.08$ N/m
2. Using volume conservation: $R^3 = n \times r^3$, so $r^3 = \frac{R^3}{512} = \frac{R^3}{8^3}$, giving $r = \frac{R}{8} = 0.125 \times 10^{-3}$ m
3. Initial surface area: $A_1 = 4\pi R^2 = 4\pi \times 10^{-6}$ m²
4. Final surface area: $A_2 = 512 \times 4\pi r^2 = 512 \times 4\pi \times \frac{R^2}{64} = 32\pi \times 10^{-6}$ m²
5. Change in surface area: $\Delta A = A_2 - A_1 = 28\pi \times 10^{-6}$ m²
6. Change in surface energy: $\Delta E = T \times \Delta A = 0.08 \times 28\pi \times 10^{-6} = 2.24\pi \times 10^{-6} \approx 7.04 \times 10^{-6}$ J
7. Therefore, $\alpha = 7$

Answer: 36

Key idea: Single Slit Diffraction - Angular Width of Central Maximum

1. For single slit diffraction, first minimum occurs at $a \sin \theta = \lambda$, where $a$ is slit width
2. For small angles, $\theta \approx \lambda/a = \frac{628 \times 10^{-9}}{2 \times 10^{-4}} = 3.14 \times 10^{-3}$ radians
3. Angular width of central maximum = $2\theta = 2 \times 3.14 \times 10^{-3} = 6.28 \times 10^{-3}$ radians
4. Convert to degrees: $6.28 \times 10^{-3} \times \frac{180}{\pi} = 6.28 \times 10^{-3} \times 57.296$
5. Angular width = $0.35978$ degrees = $35.978 \times 10^{-2}$ degrees
6. Therefore, $\alpha = 36$

Answer: 120

Key idea: Adiabatic Process and Work Done

1. Calculate $\gamma = \frac{c_p}{c_v} = \frac{3R}{2R} = 1.5$
2. For adiabatic process, find final temperature: $\frac{T_2}{T_1} = \left(\frac{P_2}{P_1}\right)^{\frac{\gamma-1}{\gamma}} = \left(\frac{1}{8}\right)^{\frac{1}{3}} = \frac{1}{2}$, so $T_2 = 206.5$ K
3. Find final volume using $P_1V_1^\gamma = P_2V_2^\gamma$: $V_2 = 0.6$ m³
4. Apply work formula for adiabatic process: $W = \frac{P_1V_1 - P_2V_2}{\gamma - 1}$
5. Calculate: $W = \frac{8 \times 10^5 \times 0.15 - 1 \times 10^5 \times 0.6}{0.5} = \frac{60000}{0.5} = 120000$ J = 120 kJ

Answer: 171

Key idea: Lorentz Force and Cross Product of Vectors

1. The magnetic force on a charged particle is $\vec{F} = q(\vec{v} \times \vec{B})$ where $q = 1 \times 10^{-6}$ C
2. Calculate cross product: $\vec{v} \times \vec{B} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -2 & 3 \\ 2 & 3 & -5 \end{vmatrix}$
3. $\vec{v} \times \vec{B} = \hat{i}(10-9) - \hat{j}(-5-6) + \hat{k}(3+4) = \hat{i} + 11\hat{j} + 7\hat{k}$
4. Force vector: $\vec{F} = 10^{-6}(\hat{i} + 11\hat{j} + 7\hat{k})$ N
5. Magnitude: $|\vec{F}| = 10^{-6}\sqrt{1^2 + 11^2 + 7^2} = 10^{-6}\sqrt{171}$ N
6. Comparing with $\sqrt{\alpha} \times 10^{-6}$ N, we get $\alpha = 171$

Answer: 3

Key idea: Stress distribution in a hanging wire with non-uniform loading

1. Identify that stress at any point = (Total force below that point)/Area
2. At distance $\frac{l}{3}$ from top, remaining length below = $l - \frac{l}{3} = \frac{2l}{3}$
3. Weight per unit length of wire = $\frac{w}{l}$, so weight of wire below = $\frac{w}{l} \times \frac{2l}{3} = \frac{2w}{3}$
4. Total force at this point = $W + \frac{2w}{3}$ (weight at bottom + wire below)
5. Stress = $\frac{W + \frac{2w}{3}}{A} = \frac{W}{A} + \frac{2w}{3A} = \frac{W}{A} + \frac{2}{3}\frac{w}{A}$
6. Comparing with given expression $\frac{W}{A} + \frac{2}{\gamma}\frac{w}{A}$, we get $\frac{2}{\gamma} = \frac{2}{3}$
7. Therefore, $\gamma = 3$

Answer: 387

Key idea: Archimedes' Principle and Floatation

1. Find mass of wooden cube: $m_{wood} = 10 \times 10 \times 10 \times 0.4 = 400$ g
2. For floating cube without coin, apply equilibrium: $m_{wood} \cdot g = \rho_{water} \cdot g \cdot A \cdot h_1$ where $A = 100$ cm²
3. Calculate initial submerged depth: $400 = 1 \times 100 \times h_1 \Rightarrow h_1 = 4$ cm
4. After placing coin, new submerged depth: $h_2 = 4 + 3.87 = 7.87$ cm
5. Apply equilibrium with coin: $(m_{wood} + m_{coin}) \cdot g = \rho_{water} \cdot g \cdot A \cdot h_2$
6. Solve for coin mass: $400 + m_{coin} = 1 \times 100 \times 7.87 = 787$ g
7. Therefore, $m_{coin} = 787 - 400 = 387$ g

Answer: Option (4)

Key idea: Stoichiometry and mole concept

1. Write the balanced equation: $3\text{Fe} + 4\text{H}_2\text{O} \rightarrow \text{Fe}_3\text{O}_4 + 4\text{H}_2$
2. Calculate molar mass of steam: $\text{H}_2\text{O} = 2(1) + 16 = 18$ g/mol
3. Find moles of steam: $n_{\text{H}_2\text{O}} = \frac{18}{18} = 1$ mol
4. Use stoichiometry: From equation, 4 mol $\text{H}_2\text{O}$ reacts with 3 mol Fe, so 1 mol $\text{H}_2\text{O}$ reacts with $\frac{3}{4}$ mol Fe
5. Calculate mass of Fe: Mass $= 0.75 \times 56 = 42$ g

Answer: Option (3)

Key idea: Ionization energy of hydrogen-like ions (one-electron species)

1. Identify that $Li^{2+}$ is a hydrogen-like ion (one electron) with atomic number $Z = 3$
2. Use the formula for ionization energy of hydrogen-like ions: $E = Z^2 \times E_H$
3. Substitute values: $E = (3)^2 \times 2.18 \times 10^{-18}$ J atom$^{-1}$
4. Calculate: $E = 9 \times 2.18 \times 10^{-18} = 19.62 \times 10^{-18}$ J atom$^{-1}$
5. Convert to proper scientific notation: $E = 1.962 \times 10^{-17}$ J atom$^{-1}$

Answer: Option (3)

Key idea: Molecular Orbital Theory - Bond Order and Magnetic Properties

1. Calculate bond order for each species using: Bond Order = $\frac{\text{bonding e}^- - \text{antibonding e}^-}{2}$
2. For $O_2^+$ (15e⁻): BO = 2.5; $O_2$ (16e⁻): BO = 2; $O_2^-$ (17e⁻): BO = 1.5; $O_2^{2-}$ (18e⁻): BO = 1
3. Since bond length ∝ $\frac{1}{\text{Bond Order}}$, Statement I gives correct sequence: $O_2^+ < O_2 < O_2^- < O_2^{2-}$ ✓
4. Count unpaired electrons from MO configuration: $O_2$ has 2 (π*²), $O_2^+$ has 1 (π*¹), $O_2^-$ has 1 (π*³), $O_2^{2-}$ has 0 (π*⁴)
5. Statement II is incorrect because $O_2^+$ and $O_2^-$ both have 1 unpaired electron (not $O_2^+ > O_2^-$)
6. Therefore, Statement I is true but Statement II is false

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Answer: Option (4)

Key idea: Colligative Properties and Weak Acid Dissociation

1. Calculate moles of fluoroacetic acid: $n = \frac{19.5}{78} = 0.25$ mol
2. Calculate molality: $m = \frac{0.25}{0.5} = 0.5$ mol/kg
3. Find van't Hoff factor using $\Delta T_f = i \times K_f \times m$: $1 = i \times 1.86 \times 0.5$, so $i = 1.0753$
4. For weak acid dissociation $HA \rightleftharpoons H^+ + A^-$, we have $i = 1 + \alpha$, so $\alpha = 0.0753$
5. Calculate $K_a$ using $K_a = \frac{\alpha^2 C}{1-\alpha} = \frac{(0.0753)^2 \times 0.5}{1-0.0753} = 3 \times 10^{-3}$

Answer: Option (4)

Key idea: Solubility Product (Ksp) and Solubility Calculations

1. For Ag$_2$CrO$_4$: Write dissolution as Ag$_2$CrO$_4$(s) $\rightleftharpoons$ 2Ag$^+$ + CrO$_4^{2-}$. If solubility = $s_1$, then K$_{sp}$ = $(2s_1)^2(s_1)$ = $4s_1^3$
2. From K$_{sp}$ = 32$x$ = $4s_1^3$, we get $s_1^3$ = 8$x$, so $s_1$ = 2$\sqrt[3]{x}$
3. For AgBr: Write dissolution as AgBr(s) $\rightleftharpoons$ Ag$^+$ + Br$^-$. If solubility = $s_2$, then K$_{sp}$ = $s_2^2$
4. From K$_{sp}$ = 4$y$ = $s_2^2$, we get $s_2$ = 2$\sqrt{y}$
5. Calculate the ratio: $\frac{s_1}{s_2}$ = $\frac{2\sqrt[3]{x}}{2\sqrt{y}}$ = $\frac{\sqrt[3]{x}}{\sqrt{y}}$

Answer: Option (1)

Key idea: Electrochemical cells, cell potential calculation, identifying cathode and anode, writing overall cell reactions

1. Identify the half-cell with higher $E°$ as cathode: $AgBr(s) + e^- \rightarrow Ag(s) + Br^-(aq)$, $E° = +0.07$ V
2. The other half-cell acts as anode (oxidation): $Fe(s) + 2OH^-(aq) \rightarrow Fe(OH)_2(s) + 2e^-$, $E° = -0.88$ V
3. Balance electrons by multiplying cathode reaction by 2: $2AgBr(s) + 2e^- \rightarrow 2Ag(s) + 2Br^-(aq)$
4. Add both half-reactions to get overall reaction: $Fe(s) + 2OH^-(aq) + 2AgBr(s) \rightarrow Fe(OH)_2(s) + 2Ag(s) + 2Br^-(aq)$
5. Calculate $E_{cell}^0 = E_{cathode}^0 - E_{anode}^0 = 0.07 - (-0.88) = +0.95$ V
6. Verify: Fe is oxidized (not reduced), $E_{cell}^0$ is intensive property, so option (1) is correct

Answer: Option (2)

Key idea: Ionization Enthalpy Trends and Electronic Configuration

1. **Statement I Analysis**: Check first ionization enthalpy trend for Na, Mg, Cl, Ar
2. Electronic configurations: Na=[Ne]3s¹, Mg=[Ne]3s², Cl=[Ne]3s²3p⁵, Ar=[Ne]3s²3p⁶
3. Correct order: Ar > Cl > Mg > Na (noble gas highest, then increases across period). Statement I claims Na > Mg > Cl > Ar, which is **FALSE**
4. **Statement II Analysis**: Check third IE for Ca, Al, Fe, B
5. Ca: [Ar]4s² → After 2 electrons removed, Ca²⁺=[Ar]. Third IE breaks stable noble gas core, so IE₃(Ca) is **very high**. Statement II is **TRUE**
6. Since Statement I is false and Statement II is true, but only two options given, answer is **(2)** as Statement I is definitely false

Answer: Option (1)

Key idea: Oxidizing power of halogens and Layer test for halide identification

1. Analyze Statement I: Oxidizing power of halogens decreases as $F_2 > Cl_2 > Br_2 > I_2$ based on reduction potentials (+2.87V, +1.36V, +1.07V, +0.54V respectively)
2. This order is indeed the basis of layer test since $Cl_2$ can oxidize $Br^-$ and $I^-$ due to higher oxidizing power. Statement I is TRUE
3. Analyze Statement II: Layer test involves reactions like $Cl_2 + 2Br^- \rightarrow 2Cl^- + Br_2$ and $Cl_2 + 2I^- \rightarrow 2Cl^- + I_2$
4. These are displacement redox reactions where $Cl_2$ displaces $Br_2$ or $I_2$ from their salts. Statement II is TRUE
5. The liberated $Br_2$ (orange-brown) or $I_2$ (violet) is extracted in organic layer for identification
6. Since both statements are correct, the answer is option (1)

Answer: Option (1)

Key idea: Redox reactions with dichromate ion; identifying reducing agents that can reduce Cr(VI) to Cr(III)

1. $K_2Cr_2O_7$ is orange due to $Cr_2O_7^{2-}$ with Cr in +6 oxidation state. In acidic medium, it acts as a strong oxidizing agent.
2. When reduced, $Cr^{6+} \to Cr^{3+}$, changing color from orange to green. We need species that can act as reducing agents.
3. Analyze each option: Reducing agents include $Fe^{2+}$, $Sn^{2+}$, $I^-$, $S^{2-}$, $NO_2^-$, $SO_2$, $C_2O_4^{2-}$ (can be oxidized).
4. Non-reducing agents: $Fe^{3+}$, $Sn^{4+}$, $SO_4^{2-}$ (already in high/highest oxidation states).
5. Option (1): $Fe^{2+}, Sn^{2+}, I^-, S^{2-}$ - All are reducing agents that can reduce dichromate.
6. Options (2), (3), (4) contain species like $Fe^{3+}$, $Sn^{4+}$, or $SO_4^{2-}$ which cannot reduce dichromate.

Answer: Option (4)

Key idea: Crystal Field Splitting Energy and Spectrochemical Series

1. Recall the spectrochemical series for ligand field strength: $CN^- > en > H_2O > F^-$
2. Stronger field ligands produce larger crystal field splitting energy ($\Delta_0$)
3. $[Cr(CN)_6]^{3-}$: CN⁻ is strongest field ligand → highest $\Delta_0 = 26,600$ cm⁻¹ (IV)
4. $[CrF_6]^{3-}$: F⁻ is weakest field ligand → lowest $\Delta_0 = 15,060$ cm⁻¹ (I)
5. $[Cr(H_2O)_6]^{3+}$: H₂O is moderate field → $\Delta_0 = 17,400$ cm⁻¹ (II)
6. $[Cr(en)_3]^{3+}$: en is strong field (between CN⁻ and H₂O) → $\Delta_0 = 22,300$ cm⁻¹ (III)
7. Match: A-IV, B-I, C-II, D-III → Option (4)

Answer: Option (1)

Key idea: Separation techniques: Simple distillation vs azeotropic mixtures

1. **Statement I Analysis:** 1,2,3-Trihydroxypropane is glycerol with b.p. = 290°C
2. Water has b.p. = 100°C. The large difference (190°C) allows separation by simple distillation
3. Glycerol and water don't form an azeotrope, so **Statement I is TRUE**
4. **Statement II Analysis:** Azeotropic mixtures have constant boiling composition where vapor and liquid have same composition
5. Since vapor and liquid compositions are identical in azeotropes, fractional distillation cannot separate them. **Statement II is TRUE**
6. Both statements are true, so answer is option **(1)**

Answer: 1

Key idea: SN1 mechanism and carbocation stability - Resonance vs Hyperconjugation

1. In $S_N1$ mechanism, the rate-determining step is carbocation formation, so reaction rate depends on carbocation stability
2. Benzyl chloride ($C_6H_5CH_2Cl$) forms benzyl carbocation ($C_6H_5CH_2^+$) which is stabilized by resonance with the benzene ring
3. Ethyl chloride ($CH_3CH_2Cl$) forms ethyl carbocation ($CH_3CH_2^+$), a primary carbocation stabilized only by weak hyperconjugation
4. Resonance stabilization is much stronger than hyperconjugation, making benzyl carbocation more stable than ethyl carbocation
5. Since benzyl carbocation is more stable, benzyl chloride reacts faster in $S_N1$ mechanism - Statement I is TRUE
6. Statement II correctly explains that ethyl carbocation (hyperconjugation) is less stabilized than benzyl carbocation (resonance) - Statement II is TRUE
7. Both statements are true and Statement II correctly explains Statement I

Answer: Option (3)

Key idea: IUPAC Priority Order of Functional Groups

1. Recall the IUPAC priority order: Carboxylic acid > Ester > Acid halide > Amide > Nitrile > Aldehyde > Ketone > Alcohol > Amine > Alkene > Alkyne
2. Check option (1): $-CONH_2, \times C=O, -CHO, -NH_2, -C\equiv C-$ - Incorrect because ketone should come after aldehyde
3. Check option (2): $-CONH_2, -COOCH_3, -CHO, -NH_2, -OH$ - Incorrect because alcohol ($-OH$) has higher priority than amine ($-NH_2$)
4. Check option (3): $-CONH_2, -CHO, \times C=O, -NH_2, -C\equiv C-$ - This follows correct order: Amide > Aldehyde > Ketone > Amine > Alkyne
5. Check option (4): $-CONH_2, -CHO, -CN, -NH_2, -C\equiv C-$ - Incorrect because nitrile ($-CN$) has higher priority than aldehyde ($-CHO$)
6. Option (3) is the correct answer

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Key idea: Electrophilic Aromatic Substitution - Activating and Deactivating Groups in Amides

Answer: Option (4)

Key idea: Distinguishing tests for organic compounds - Carbylamine test, Tollen's test, FeCl₃ test, and Bromine water test

1. **A. Diethyl amine + Ethyl amine**: Use Carbylamine test ($CHCl_3 + KOH, \Delta$). Primary amine (ethyl amine) gives foul-smelling isocyanide, secondary amine (diethyl amine) doesn't. So A → II
2. **B. Acetaldehyde + Acetone**: Use Tollen's reagent (Ammonical silver nitrate). Aldehyde gives silver mirror test (positive), ketone doesn't. So B → IV
3. **C. Ethanol + Phenol**: Use Neutral $FeCl_3$. Phenol gives violet coloration, ethanol doesn't react. So C → III
4. **D. Benzoic acid + Cinnamic acid**: Use Bromine water. Cinnamic acid has $C=C$ double bond and decolorizes bromine water, benzoic acid (saturated) doesn't. So D → I
5. Therefore, the correct matching is: A-II, B-IV, C-III, D-I

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Key idea: Azo coupling reaction - electrophilic aromatic substitution with diazonium salts, considering both electronic activation and steric accessibility

Answer: Option (3)

Key idea: Vitamins and their chemical names - specifically water-soluble vitamins (B-complex and C)

1. Identify that Vitamin $B_1$ is **Thiamine**, so A → III
2. Identify that Vitamin $B_2$ is **Riboflavin**, so B → IV
3. Identify that Vitamin $B_6$ is **Pyridoxine**, so C → I
4. Identify that Vitamin C is **Ascorbic acid**, so D → II
5. Match the pattern: A-III, B-IV, C-I, D-II corresponds to option (3)

Answer: Option (2)

Key idea: Qualitative Analysis - Cation and Anion Identification through Flame Test, Group Precipitation, and Specific Anion Tests

1. **Flame Test Analysis:** Apple green flame with conc. HCl indicates $Ba^{2+}$ cation (characteristic color for barium)
2. **Confirmation of Cation:** Group precipitate + acetic acid + $K_2CrO_4$ gives yellow precipitate of $BaCrO_4$, confirming $Ba^{2+}$
3. **Anion Test:** Sodium carbonate extract converts the salt to soluble form for anion testing
4. **Phosphate Test:** Heating with conc. $HNO_3$ + ammonium molybdate gives canary yellow precipitate of $(NH_4)_3[PO_4 \cdot 12MoO_3]$
5. **Conclusion:** The canary yellow precipitate is specific for $PO_4^{3-}$ anion. Therefore, cation = $Ba^{2+}$ and anion = $PO_4^{3-}$

Answer: 33

Key idea: Coordination compounds, electrolyte behavior, ion exchange, and stoichiometry

1. Calculate molar mass of $\text{CrCl}_3 \cdot 6\text{H}_2\text{O} = 52 + 3(35.5) + 6(18) = 266.5$ g/mol
2. Find moles of complex: $n_{\text{complex}} = \frac{5.33}{266.5} = 0.02$ mol
3. Since it's a 1:3 electrolyte, the structure is $[\text{Cr}(\text{H}_2\text{O})_6]\text{Cl}_3$ which gives 3 Cl$^-$ ions per complex
4. Calculate moles of AgCl formed: Molar mass of AgCl $= 143.5$ g/mol, so $n_{\text{AgCl}} = \frac{8.61}{143.5} = 0.06$ mol
5. Find the ratio: $\frac{n_{\text{complex}}}{n_{\text{AgCl}}} = \frac{0.02}{0.06} = \frac{1}{3}$
6. Convert to required format: $\frac{1}{3} \times 10^2 = 33.33 \approx 33$

Answer: 14

Key idea: Free radical halogenation of cycloalkanes and identification of unique monochloro products based on molecular symmetry

1. Identify that $\text{C}_5\text{H}_{10}$ with degree of unsaturation = 1 and no $\text{KMnO}_4$ decolorization means cycloalkanes (not alkenes)
2. List all cycloalkane isomers: cyclopentane, methylcyclobutane, ethylcyclopropane, 1,1-dimethylcyclopropane, and 1,2-dimethylcyclopropane
3. For cyclopentane: all H equivalent → 1 monochloro product
4. For methylcyclobutane: 4 types of non-equivalent H → 4 monochloro products
5. For ethylcyclopropane: 4 types of non-equivalent H → 4 monochloro products
6. For 1,1-dimethylcyclopropane: 2 types of non-equivalent H → 2 monochloro products
7. For 1,2-dimethylcyclopropane: 3 types of non-equivalent H (by symmetry) → 3 monochloro products. Total = 1+4+4+2+3 = 14

Answer: 17

Key idea: Complete combustion of alkanes and stoichiometry

1. For an alkane $C_nH_{2n+2}$, write the combustion equation: $C_nH_{2n+2} + O_2 \to CO_2 + H_2O$
2. Balance the equation: $C_nH_{2n+2} + \frac{3n+1}{2}O_2 \to nCO_2 + (n+1)H_2O$
3. Given that moles of $O_2$ required = 8, so $\frac{3n+1}{2} = 8$
4. Solve: $3n+1 = 16 \Rightarrow n = 5$
5. The alkane is $C_5H_{12}$ with 5 carbon atoms and 12 hydrogen atoms
6. Sum of carbon and hydrogen atoms = $5 + 12 = 17$

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Key idea: Arrhenius Equation - Temperature Dependence of Rate Constant

Answer: 727

Key idea: Gibbs Free Energy at Phase Transition

1. At transition temperature, the two states A and B are in equilibrium, so $\Delta G^0 = 0$
2. Set the equation to zero: $105 - 35 \log T = 0$
3. Solve for $\log T$: $\log T = \frac{105}{35} = 3$
4. Find T in Kelvin: $T = 10^3 = 1000$ K
5. Convert to Celsius: $T(°C) = 1000 - 273 = 727°C$